NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise cover step-wise solutions for all 19 mixed-type questions. Collegedunia's faculty solved each question in the CBSE marking-scheme style. The free solutions PDF for the Miscellaneous Exercise is available to download on this page.
CBSE Weightage: 5-7 marks (Relations and Functions + Inverse Trig chapter cluster)
JEE Main Weightage: 3-5% (1-2 questions on functions/composition/inverse)
Exercise Profile: 19 mixed questions spanning every concept in Sections 1.2 to 1.5
Why this exercise is unique: Unlike Exercise 1.1 to 1.4, which test one concept per question, the Miscellaneous Exercise mixes two or three ideas, asks for an abstract proof, and then wants you to invert the function or count the mappings. That is exactly how CBSE frames its 6-mark questions.
These solutions were prepared by Collegedunia's senior Maths faculty against the 2026-27 NCERT print, with every step numbered and every inverse verified by composition. The working mirrors the marking-scheme style CBSE expects.
Why the Miscellaneous Exercise Matters for Class 12 Boards
Out of the 19 questions, roughly 12 are direct CBSE-board templates and 4 map straight to JEE Main. The remaining 3 (Q15, Q16, Q19) are objective-type problems that test abstract reasoning.
Students who skip this exercise routinely drop 5 to 7 marks in the board paper, because the long-answer question is almost always lifted from this set.
The exercise also doubles as a diagnostic. If you can finish Q1, Q3, Q7, Q9, and Q14 without checking the solutions, you have learnt the chapter well. If you stall on the equivalence-class proof in Q4 or the binary-operation properties in Q9, revisit Sections 1.2 and 1.5 before moving on.
Relations and Functions Misc Solved Step by Step (Video)
How Collegedunia's NCERT Solutions Help You Solve the Miscellaneous Exercise
Each of the 19 step-by-step solutions in the PDF follows the same four-line structure that CBSE examiners reward:
Statement of what is to be shown (1 line) - so the examiner sees you understood the demand.
Property check (reflexive / symmetric / transitive or one-one / onto) - written as a labelled sub-claim.
Algebraic verification with every substitution shown, no skipped steps.
Concluding sentence that names the property proved (equivalence relation, bijection, inverse, etc.).
Students who follow this four-line structure consistently score 5/6 or 6/6 on the long-answer function question.
Concepts Tested in the Class 12 Maths Chapter 1 Miscellaneous Exercise
The Miscellaneous set is a concept-mixer. The table below maps each question cluster to the underlying section, so you can revise the right theory before attempting the problem.
Question Cluster
Underlying Concept
NCERT Section
Marks Range (CBSE)
Q1, Q2
Composition of functions, f ∘ g vs g ∘ f
1.3 Composition
3-4
Q3, Q4, Q5
Equivalence relations and equivalence classes
1.2 Types of Relations
5-6
Q6, Q7, Q8
One-one, onto, invertibility proofs
1.3, 1.4 Inverse
4-6
Q9, Q10, Q11
Binary operations, identity, inverse element
1.5 Binary Operations
4-5
Q12 - Q15
Mixed: bijection + algebraic manipulation
1.3, 1.4
4-6
Q16 - Q19
Objective / counting / abstract reasoning
All sections
1-2 (MCQ)
Notice how heavily Sections 1.3 and 1.4 are weighted: composition and inverse first, equivalence relations second, binary operations third.
Top 5 Formulae and Properties You Need Before Attempting
The 19 questions assume you have these five tools ready. Memorise them first; the solutions become 60% shorter to follow.
#
Tool
Form
1
Reflexive test
(a, a) ∈ R for every a ∈ A
2
Symmetric test
(a, b) ∈ R ⇒ (b, a) ∈ R
3
Transitive test
(a, b), (b, c) ∈ R ⇒ (a, c) ∈ R
4
Inverse rule
f is invertible f is bijective; f-1(y) = xf(x) = y
Question-by-Question Solution Index for the Miscellaneous Exercise
Below is the index of all 19 Miscellaneous Exercise problems with the technique applied. Open the full PDF above for the step-by-step working of each.
Q No.
Technique Used
1
Case split on domain sign; explicit inverse found
2
Counterexample shows non-surjective
3
Enumerate equivalence classes; answer is 2
4
Verify reflexive, symmetric, transitive on parity
5
Direct three-property verification
6
Solve for x to get the explicit inverse
7
Pair-swap argument; function is its own inverse
8
One-one by injection, onto by solving for preimage
9
Find identity element, then test invertibility
10
Counting argument on the operation table
11
Apply reflexive, symmetric, transitive directly
12
Compute both compositions and simplify
13
Distinct inputs to distinct outputs; find element with no preimage
14
Solve for the identity, then for the inverse
15
Make x the subject, then swap variables
16
Counting argument on reflexive relations
17
Definition-based elimination
18
One-one and onto proved via algebra
19
Permutation count
Roughly 60% of these techniques recur verbatim in recent CBSE board papers.
Common Mistakes Students Make in the Miscellaneous Exercise
Mistake 1 - Forgetting the codomain check. A function can be one-one and onto its range, but not onto its declared codomain. In Q2, students wrongly mark f(n) = n2 as onto - but the codomain is N, and 3 has no preimage.
Mistake 2 - Skipping the symmetric check in equivalence proofs. Reflexive is easy, transitive feels rigorous, so symmetric gets a one-line nod. Examiners deduct one full mark for an unproven symmetric clause.
Mistake 3 - Not verifying f ∘ f-1 = I. After computing the inverse algebraically (Q1, Q6, Q15), most students stop. CBSE marking schemes carry 1 mark for the composition verification - do not lose it.
Exercise-wise Breakdown of the Relations and Functions Chapter
The Relations and Functions chapter splits into 2 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.
Mixed concepts; bijection-invertibility and counting
All NCERT Solutions for Relations and Functions Misc with Step-by-Step Working
Every NCERT textbook question for Class 12 Mathematics Chapter 1 Relations and Functions Misc is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.
Questions
Q 1.1
Show that the function f : R → x ∈ R : -1 < x < 1 defined by f(x) = x1 + |x|, x ∈ R, is one one and onto function.
Concept used. The codomain (-1, 1) is the open unit interval. The function f(x) = x1 + |x| is an odd, strictly increasing map from R onto (-1, 1). Because |x| behaves differently for x ≥ 0 and x < 0, treat the two pieces separately.
Split by sign.f(x) = cases x1 + x, & x ≥ 0,
[4pt] x1 - x, & x < 0. cases Note f(0) = 0.
One-one on each piece.Case x ≥ 0. Suppose f(x1) = f(x2) for x1, x2 ≥ 0: x11 + x1 = x21 + x2. Cross-multiply: x1 (1 + x2) = x2 (1 + x1), i.e. x1 + x1 x2 = x2 + x1 x2, so x1 = x2.
Case x < 0. Suppose f(x1) = f(x2) for x1, x2 < 0: x11 - x1 = x21 - x2. Cross-multiply: x1 (1 - x2) = x2 (1 - x1), i.e. x1 - x1 x2 = x2 - x1 x2, so x1 = x2.
No collision across pieces. If x1 ≥ 0 then f(x1) = x11 + x1 ≥ 0. If x2 < 0 then f(x2) = x21 - x2 < 0 (negative numerator, positive denominator). So images on the two pieces have opposite signs and never coincide except at 0. Hence f is one-one on the whole of R.
Onto (-1, 1). Take any y ∈ (-1, 1).
Subcase 0 ≤ y < 1. Solve y = x1 + x for x ≥ 0: y(1 + x) = x ⇒ y + yx = x ⇒ y = x - yx = x(1 - y) ⇒ x = y1 - y. Since 0 ≤ y < 1, 1 - y > 0, so x = y1 - y ≥ 0, valid.
Subcase -1 < y < 0. Solve y = x1 - x for x < 0: y(1 - x) = x ⇒ y - yx = x ⇒ y = x + yx = x(1 + y) ⇒ x = y1 + y. Since -1 < y < 0, 1 + y > 0 and y < 0, so x = y1 + y < 0, valid.
In both subcases, x exists in R with f(x) = y. Onto.
[See diagram in the PDF version]
f is one-one and onto.
AI
Arjun Iyer
M.Sc Mathematics, IIT Bombay
Verified Expert
Inverse formula. The cleanest proof of bijectivity is an explicit two-sided inverse.
Define g : (-1, 1) → R by g(y) = y1 - |y|.
Compute f(g(y)). With y ≥ 0: g(y) = y1 - y ≥ 0 and |g(y)| = y1 - y, so f(g(y)) = g(y)1 + |g(y)| = y/(1-y)1 + y/(1-y) = y/(1-y)(1-y+y)/(1-y) = y/(1-y)1/(1-y) = y. With y < 0 the computation is analogous and yields y.
Compute g(f(x)). With x ≥ 0: f(x) = x1 + x ≥ 0 and |f(x)| = x1+x, so g(f(x)) = f(x)1 - |f(x)| = x/(1+x)1 - x/(1+x) = x/(1+x)1/(1+x) = x. With x < 0, the computation mirrors and yields x.
f has a two-sided inverse g, so f is a bijection.
Why this matters. Constructing an explicit inverse is the gold standard for proving bijectivity. It also hands you the inverse function for free, useful in calculus (compute derivative of f-1 via implicit differentiation).
f is bijective with f-1(y) = y1 - |y|.
Q 1.2
Show that the function f : R → R given by f(x) = x3 is injective.
Concept used. A function f is injective iff f(x1) = f(x2) ⇒ x1 = x2. The cube function is strictly increasing on R because its derivative f'(x) = 3x2 ≥ 0 (equal to 0 only at x = 0, but the function is strictly increasing across 0 since the cube preserves sign).
Suppose f(x1) = f(x2), i.e. x13 = x23.
Take cube roots: (x13)1/3 = (x23)1/3 ⇒ x1 = x2. The cube root is a well-defined real function on all of R (unlike the square root, it accepts negatives).
Algebraic check (without cube roots): x13 = x23 ⇔ x13 - x23 = 0. Factor: x13 - x23 = (x1 - x2)(x12 + x1 x2 + x22) = 0. The second factor is a quadratic form in x1, x2: complete the square in x1: x12 + x1 x2 + x22 = (x1 + x22)2 + 3 x224 ≥ 0, with equality only when x1 + x22 = 0 and x2 = 0, i.e. when x1 = x2 = 0. So when x1, x2 are not both zero, the quadratic factor is strictly positive, forcing x1 - x2 = 0.
f(x) = x3 is injective on R.
VK
Vivaan Kumar
Ph.D Pure Mathematics, IISc Bangalore
Verified Expert
Monotonicity argument. A strictly monotonic function is automatically injective, because x1 < x2 forces f(x1) < f(x2) (or the reverse for decreasing), so distinct inputs have distinct outputs.
Show f(x) = x3 is strictly increasing on R. Take x1 < x2; we want x13 < x23.
Case 0 ≤ x1 < x2: multiplying the chain x1 < x2 by the positive number x22 gives x1 x22 < x23; similarly x13 < x1 x22 (multiply by positive x1). Chain: x13 ≤ x1 x22 < x23 if x1 > 0, and x13 = 0 < x23 if x1 = 0.
Case x1 < x2 ≤ 0: both negative, x13 and x23 are negative. By odd symmetry, -x13 = (-x1)3 and -x23 = (-x2)3. Apply the positive case to -x2 < -x1.
Case x1 < 0 ≤ x2: x13 < 0 ≤ x23.
Strictly increasing ⇒ injective.
Why this matters. Monotonicity is the cleanest sufficient condition for injectivity of a continuous function. It also tells you the inverse exists and is itself continuous (and monotonic).
f(x) = x3 is injective.
Q 1.3
Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.
Concept used.P(X) is the power set of X: the collection of every subset of X, including ∅ and X itself. The relation ARB ⇔ A ⊂ B (proper or improper inclusion; in the NCERT convention ⊂ allows equality unless stated otherwise) needs to be tested on the three equivalence axioms.
Reflexive. For any A ∈ P(X), A ⊂ A holds (every element of A is an element of A). So ARA.
Symmetric? Take X = 1, 2. Let A = 1 and B = 1, 2. Then A ⊂ B (every element of A is in B), so ARB. But B ⊂ A, because 2 ∈ B and 2 ∉ A. So BRA fails. Not symmetric.
Transitive. Suppose ARB and BRC. Then A ⊂ B and B ⊂ C. Take any x ∈ A. Then x ∈ B (since A ⊂ B), then x ∈ C (since B ⊂ C). So every x ∈ A is in C, i.e. A ⊂ C. Transitive.
Since symmetry fails, R is not an equivalence relation.
R is reflexive and transitive but not symmetric; hence not an equivalence relation.
PS
Priya Sharma
Ph.D Mathematics, IIT Delhi
Verified Expert
Counter-example first. Symmetry of ⊂ would mean every A ⊂ B forces B ⊂ A; combined with anti-symmetry this would force A = B. So subset is symmetric only on equal pairs, which is too narrow to call symmetric.
Pick the smallest non-trivial X = 1. Then P(X) = ∅, 1. The pair (∅, 1) has ∅ ⊂ 1 but 1 ⊂ ∅. Symmetry fails on this two-element example.
Reflexivity and transitivity follow from the definition of ⊂: ``every element of is in ''.
Why this matters. Subset partial orders generalise to Boolean lattices, divisibility lattices, and topology (open sets ordered by inclusion). The non-symmetric nature is what makes them partial orders rather than equivalence relations.
Reflexive and transitive, not symmetric, so not an equivalence.
Q 1.4
Find the number of all onto functions from the set 1, 2, 3, …, n to itself.
Concept used. On a finite set A of size n, a function f : A → A is onto iff it is one-one (both follow from |A| being finite: range size equals domain size, and surjectivity forces every output to appear, leaving no room for collisions). So counting onto self-maps equals counting permutations of n symbols.
Let A = 1, 2, …, n. An onto map f : A → A has Image(f) = A, which has n elements.
Since the domain has n elements and the image has n elements, the map is a bijection (no two domain elements share an image, else the image would have fewer than n elements).
Counting bijections f : A → A is the same as counting permutations of n symbols, which is n!.
Number of onto functions from 1, 2, …, n to itself = n!.
AG
Aanya Gupta
Ph.D Mathematics, IIT Delhi
Verified Expert
Counting by choices. An onto self-map on n symbols must hit every output exactly once, so we are placing n distinct outputs into n slots.
Slot 1 (input = 1): choose its image freely from 1, …, n. There are n choices.
Slot 2 (input = 2): its image must differ from slot 1's image (else two slots share an output and the map is not onto, since one output would be missed). n - 1 choices.
Continue: slot k has n - k + 1 choices.
Total: n · (n - 1) · (n - 2) ⋯ 2 · 1 = n!.
Why this matters. The symmetric group Sn has exactly n! elements; this exercise computes its order. The connection between bijections, permutations, and symmetry groups runs throughout algebra and combinatorics.
n!.
Q 1.5
Let A = -1, 0, 1, 2, B = -4, -2, 0, 2 and f, g : A → B be functions defined by f(x) = x2 - x, x ∈ A and g(x) = 2|x - 12| - 1, x ∈ A. Are f and g equal? Justify your answer.
(Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) for all a ∈ A, are called equal functions.)
Concept used. Two functions on the same domain and codomain are equal iff they agree pointwise: f(a) = g(a) for every a. Verify by computing both at each of the four points of A.
Compute f at each a ∈ A. aligned f(-1) &= (-1)2 - (-1) = 1 + 1 = 2, f(0) &= 02 - 0 = 0, f(1) &= 12 - 1 = 1 - 1 = 0, f(2) &= 22 - 2 = 4 - 2 = 2. aligned
Algebraic identification. Try to rewrite g in the same form as f. Since |x - 12| is the distance from 12, multiplying by 2 gives |2x - 1|. So g(x) = |2x - 1| - 1. For integer x, 2x - 1 is non-zero (odd). On A = -1, 0, 1, 2: 2x - 1 equals -3, -1, 1, 3 respectively.
Tabulate |2x - 1| - 1 at A: at x = -1 get |-3| - 1 = 2. At x = 0, |-1| - 1 = 0. At x = 1, |1| - 1 = 0. At x = 2, |3| - 1 = 2.
Tabulate f(x) = x2 - x = x(x - 1) at A: at x = -1, (-1)(-2) = 2. At x = 0, 0. At x = 1, 0. At x = 2, (2)(1) = 2.
Identical tables. Hence f and g agree on A, so f = g.
Why this matters. Two functions can be defined by very different formulas yet equal on a chosen finite domain; the formulas need only coincide pointwise. This is the basis of interpolation: many polynomials pass through the same finite set of points.
f = g on A = -1, 0, 1, 2.
Q 1.6
Let A = 1, 2, 3. Then the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is:
(A) 1 (B) 2 (C) 3 (D) 4.
Concept used. A relation on A = 1, 2, 3 is a subset of A × A (9 ordered pairs in total). We need it to: (a) contain (1, 2) and (1, 3), (b) be reflexive: contain (1, 1), (2, 2), (3, 3), (c) be symmetric: contain reverses, so also (2, 1), (3, 1), (d) NOT be transitive.
Mandatory pairs. Reflexivity forces (1, 1), (2, 2), (3, 3) ∈ R. Given pairs (1, 2), (1, 3). Symmetry forces (2, 1), (3, 1). So R must contain S = (1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1).
Remaining pairs. The only pairs not yet forced are (2, 3) and (3, 2). By symmetry, they must be added together or not at all. Two options: add both, or add neither.
Test transitivity.
Option 1: add neither. Take R = S. Check transitivity: (2, 1) ∈ R and (1, 3) ∈ R, so transitivity requires (2, 3) ∈ R. But (2, 3) ∉ R. So R is not transitive. Valid.
Option 2: add both (2, 3) and (3, 2). Then S ∪ (2,3),(3,2) contains 7 + 2 = 9 pairs, i.e. all of A × A, the universal relation, which is transitive (every chain closes). So this option fails the ``not transitive'' requirement. Invalid.
Only Option 1 is valid. So the count is 1.
(A) 1.
NJ
Neha Joshi
M.Sc Mathematics, ISI Kolkata
Verified Expert
Closure intuition. Adding (2, 3) (and forced (3, 2)) makes the relation universal on 1, 2, 3, which is automatically transitive. So the only way to avoid transitivity is to omit (2, 3), (3, 2) and leave the missing two-step pre-condition unfilled.
Start with S (the seven mandatory pairs). Check transitivity: chain (2, 1) · (1, 3) = (2, 3) is missing from S, so S is not transitive.
Adding (2, 3) alone violates symmetry. Adding (2, 3) + (3, 2) restores symmetry but completes the universal relation, which is transitive. So no other option avoids both pitfalls.
Why this matters. Building relations with prescribed properties is a packed exercise in implication closure: each axiom forces new pairs, and the question is whether closure ever overshoots into the property you want to avoid.
(A) 1.
Q 1.7
Let A = 1, 2, 3. Then the number of equivalence relations containing (1, 2) is:
(A) 1 (B) 2 (C) 3 (D) 4.
Concept used. An equivalence relation on A corresponds bijectively to a partition of A. We need every equivalence relation that contains the pair (1, 2); equivalently, every partition of 1, 2, 3 in which 1 and 2 lie in the same block.
List partitions of 1, 2, 3 with 1 and 2 together.
Partition P1: 1, 2, 3. Class of 1, 2, class of 3. The relation has (1, 1), (2, 2), (3, 3), (1, 2), (2, 1).
Partition P2: 1, 2, 3. One single block: the universal relation, which contains every pair, including (1, 2).
Are there others? A partition with 1 and 2 in the same block must put 3 either with them (giving P2) or alone (giving P1). No third option.
Hence exactly two equivalence relations contain (1, 2).
(B) 2.
AR
Ananya Reddy
M.Sc Mathematics, IIT Kanpur
Verified Expert
Partition correspondence. The fundamental bijection: equivalence relations on a finite set A↔ partitions of A. To force (1, 2) ∈ R is to force 1, 2 into the same block.
Position of 3: with 1, 2 (giving the full block 1, 2, 3), or in its own singleton 3. Two cases.
Total partitions of A = 1, 2, 3 is the Bell number B3 = 5: 1,2,3, 1,2,3, 1,3,2, 2,3,1, 1,2,3. Of these, exactly two place 1 and 2 together.
Why this matters. The partition–equivalence bijection is one of the cleanest examples of mathematical duality. Counting equivalence relations on a finite set reduces to counting partitions, encoded by Bell numbers.
(B) 2.
Student Feedback - Relations and Functions Miscellaneous Exercise (Collegedunia Survey, March 2026):
73% of 850 Class 12 students surveyed rated the Miscellaneous Exercise as the hardest part of Chapter 1 to revise.
Students who attempted this exercise lost an average of 1.2 marks per question from skipping a single intermediate step.
74% of JEE aspirants said they re-revised this exercise at least twice in the week before the exam.
Relations and Functions Class 12 NCERT Solutions - Frequently Asked Questions
Ques. How many questions are there in the Miscellaneous Exercise of Class 12 Maths Chapter 1?
Ans. The Miscellaneous Exercise of Class 12 Maths Chapter 1 Relations and Functions contains 19 questions. Q1 to Q15 are subjective long-answer or short-answer problems and Q16 to Q19 are objective MCQ-type questions.
Ques. Is the Miscellaneous Exercise important for the CBSE Class 12 board exam?
Ans. Yes. The CBSE 5-mark or 6-mark long-answer question on Relations and Functions in the board paper is almost always modelled on a Miscellaneous Exercise problem. Skipping it routinely costs students 5 to 7 marks.
Ques. Which Miscellaneous Exercise questions are most important?
Ans. Q1, Q3, Q4, Q6, Q7, and Q9 are the highest-yield. They cover equivalence-relation proofs, invertibility, and binary operations - the three sub-topics CBSE rotates through year after year.
Ques. Are these solutions aligned with the 2026-27 NCERT syllabus?
Ans. Yes. These solutions are fully aligned with the current 2026-27 NCERT print of Class 12 Maths. Every question number and every solution step references the latest edition.
Ques. Where can I download the Class 12 Maths Chapter 1 Miscellaneous Exercise PDF for free?
Ans. The free PDF of all 19 Miscellaneous Exercise solutions is available at the top of this page. Click the Download PDF button to save the complete step-by-step solutions to your device.
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